JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Solutions

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2 7^{\circ} C

, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is ______ g

mol^{- 1}

(Nearest integer)

(Given:

R = 0.083 L

bar

K^{- 1} mol^{- 1}

)

Answer: 62250

Solution

From

π = \frac{w}{M} \cdot \frac{R T}{V}

the molar mass is

M = \frac{w R T}{π V}

. Converting the pressure,

π = \frac{400}{1 0 ^{5}} = 0.004 bar

, with

T = 300 K

and

V = 0.250 L

. Then

M = \frac{2.5 \times 0.083 \times 300}{0.004 \times 0.250} = 62250 g mo l^{- 1}

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer

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